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(B) Let the two positive numbers be $a$ and $b$.Given that $x$ is the arithmetic mean,$x = \frac{a+b}{2}$,so $a+b = 2x$.Given that $y$ and $z$ are the

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$2$

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(B) Let the two positive numbers be $a$ and $b$.Given that $x$ is the arithmetic mean,$x = \frac{a+b}{2}$,so $a+b = 2x$.Given that $y$ and $z$ are the two geometric means between $a$ and $b$,the sequence $a, y, z, b$ is in geometric progression.Let the common ratio be $r$. Then $y = ar$,$z = ar^2$,and $b = ar^3$.Thus,$r^3 = \frac{b}{a}$,which implies $r = (\frac{b}{a})^{1/3}$.Also,$y = a(\frac{b}{a})^{1/3} = a^{2/3}b^{1/3}$ and $z = a(\frac{b}{a})^{2/3} = a^{1/3}b^{2/3}$.Then $xyz = (a^{2/3}b^{1/3})(a^{1/3}b^{2/3})x = abx$.Since $a+b = 2x$,we have $y^3 = a^2b$ and $z^3 = ab^2$.Therefore,$\frac{y^3 + z^3}{xyz} = \frac{a^2b + ab^2}{abx} = \frac{ab(a+b)}{abx} = \frac{a+b}{x} = \frac{2x}{x} = 2$.

If $x$ is the arithmetic mean and $y, z$ are the two geometric means between two positive numbers,then $\frac{y^3 + z^3}{xyz} = \dots..$

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